No and this is a very common misconception!!!
First of all, let's write $H=1$ if the alternative is true and $H=0$ if the null is true. You are asking whether the p-value is equal to $\Pr[H=0]$.
First observation: From a frequentist point of view, your question does not even make sense, because either your hypothesis is true or it is not (i.e. $\Pr[H=0]=0$ or $\Pr[H=0]=1$).
But also if you relax this a bit and take a more Bayesian point of view, so that it could be the case $0<\Pr[H=0]<1$, then the question still does not make sense, because the p-value is a random variable, while the former probability is just a constant!!!
In particular let's look again at your definition of a p-value:
"the p-value is the probability that the measurement takes a value that is at least as unlikely under the null hypothesis as the observed value"
So based on this definition (and simplifying things a bit to e.g. to one-sided tests) you observe a random variable $X$ and then the p-value is just $P=1-F_0(X)$, where $F_0$ is the distribution of $X$ under the null hypothesis. Again the key point to observe here is that the p-value is random, because $X$ is random, so it can in principle (randomly) take any values between 0 and 1. Therefore it cannot be equal to a constant ($\Pr[H=0]$)!
The other important insight is that at least for continuous $F_0$, it will hold that $P \sim U[0,1]$ under the null hypothesis. For powerful tests the alternative distributions of the p-value will be stochastically a lot smaller than uniform, i.e. $Pr[P \leq t \mid H=1] > t$. Thus if you observe a very small p-value it is a lot more likely to have come from the alternative than from a uniform distribution. Again this can be made more precise in a Bayesian framework but also explains things from a frequentist point of view.
Still the most important thing to always remember is that the p-value is a random variable.